Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
434 views
in Technique[技术] by (71.8m points)

mysql - Get results that fall within marker radiuses from database

Update 16th November 2012

I would like to raise this question again, offering with a new bounty for a solid, good solution. It seems that only the solution (shubhansh's answer) does not effectively work now. I will explain why.

Firstly, this is the live map I have with radiuses and people, the radiuses are in red and the people are in blue.

enter image description here

As you can see, there are two people in this map with eight radiuses, basically I am getting only the person which is Person A, but I am not getting Person B, I'm guessing that the SQL is not correctly picking it up which I need it to be precise and accurate from the person's radius and the marker radiuses.

It looks like what is picked up are are inside the radiuses, not those who overlap a radius, I need it to be able to pick up any results for any radiuses that overlap each other.

I am looking for a precise and accurate SQL than shubhansh's answer. You may read below to read how exactly I need the query to act and pick up accurate people.

The data, PEOPLE:

+-----------+-----------+--------+
| latitude  | longitude | radius |
+-----------+-----------+--------+
| 51.517395 | -0.053129 | 5.6    |
| 51.506607 | -0.116129 | 0.7    |
+-----------+-----------+--------+

Please note that radius is in kilometers.

+-----------+-----------+-----+
| latitude  | longitude | km  |
+-----------+-----------+-----+
| 51.502117 | -0.103340 | 0.3 |
| 51.498913 | -0.120850 | 0.7 |
| 51.496078 | -0.108919 | 0.7 |
| 51.496506 | -0.095873 | 0.7 |
| 51.503399 | -0.090723 | 0.7 |
| 51.508049 | -0.100336 | 0.7 |
| 51.508797 | -0.112610 | 0.7 |
| 51.505535 | -0.125227 | 0.7 |
| 51.502331 | -0.108061 | 0.7 |
+-----------+-----------+-----+

The current SQL I use:

SELECT ppl.latitude,
       ppl.longitude,
       ppl.radius
FROM 
(
    people ppl
),
(
    SELECT latitude, longitude 
    FROM radiuses
) AS radius
WHERE (POW((ppl.longitude - radius.longitude) * 111.12 * COS(ppl.latitude), 2) + POW((ppl.longitude - radius.longitude) * 111.12, 2)) <= 4
GROUP BY ppl.id

The data for MySQL which you can use to test your query,

INSERT INTO radiuses (id, latitude, longitude, km) VALUES ('1', '51.502117', '-0.103340', '0.3'), ('2', '51.498913', '-0.120850', '0.7'), ('3', '51.496078', '-0.108919', '0.7'), ('4', '51.496506', '-0.095873', '0.7'), ('5', '51.503399', '-0.090723', '0.7'), ('6', '51.508049', '-0.100336', '0.7'), ('7', '51.508797', '-0.112610', '0.7'), ('8', '51.505535', '-0.125227', '0.7'), ('9', '51.502331', '-0.108061', '0.7');

INSERT INTO people (id, latitude, longitude, radius) VALUES ('1', '51.517395', '-0.053129', '5.6'), ('2', '51.506607', '-0.116129', '0.7');

Old summary

Note: all the latitudes and longitudes are just randomly made.

I have a map applet which a user can place his radius of a lat/lng location, with a 1km radius.

Now, there is another user that can put his radiuses, at any location on the map, each with 1km radius (same as the user above).

Like this User A is red and User B is blue.

enter image description here

Basically User A stores his radiuses in a table that looks like this:

+-----------+---------+-----------+-----------+
| radius_id | user_id | latitude  | longitude |
+-----------+---------+-----------+-----------+
|         1 |       1 | 81.802117 | -1.110035 |
|         2 |       1 | 81.798272 | -1.144196 |
|         3 |       1 | 81.726782 | -1.135919 |
+-----------+---------+-----------+-----------+

And User B stores his radius in another table that looks like this - (note: they can only store 1 coordinates per account):

+---------+-----------+-----------+
| user_id | latitude  | longitude |
+---------+-----------+-----------+
|       6 | 81.444126 | -1.244910 |
+---------+-----------+-----------+

I want to be able to pick up those users that fall within the defined radiuses, even if the radius circles are touching, in the map picture. Only marker C would be able to pick up the single radius, when A and B do not.

I'm sure this is possible, but I do not know how to come up with this kind of system in MySQL.

I found this on the Google Developers site it's close but not just what it performs I need.

EDIT: I have found a better one, this is very close, but still not what I am looking for, since it uses 1 bound of latitude and longitude coordinates when I have multiple in a table.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

For solving this you need to understand equation of circle, which is something like this For any point (x,y) to fall within circle with center (x1, y1) and radius r units is

(x-x1)^2 + (y - y1)^2 <= r^2

where a^b = a to the power b

Here in your case User B's (latitude, longitude) are the center of circle, User A's (latitude, longitude) are the points (x,y) and radius = 2kms.

But basic problem is of changing degrees of latitudes to longitudes, so here is solution, 1 degree = 111.12 km. So to keep units same on both side of equation, we will convert it to Kms

So our final equation becomes:

((x-x1)*111.12)^2 + ((y-y1)*111.12)^2 = 4      (=2^2) 

SQL statement for the same should look something like this

SELECT A.user_id, A.radius_id, A.latitude, A.logitude
FROM UserA AS A, 
     (SELECT user_id, latitude, longitude 
       FROM UserB 
       WHERE user_id = 8) AS B
WHERE (POW((A.latitude-B.latitude)*111.12, 2) + POW((A.longitude - B.longitude)*111.12, 2)) <= 4
/* **Edit** Here I have used (A.longitude - B.longitude)*111.12, for more accurate results one can replace it with (A.longitude - B.longitude)*111.12*cos(A.latitude)) or (A.longitude - B.longitude)*111.12*cos(B.latitude)) 

And, as i have suggested in the comments that first filter some records based on approximation, so whether one uses A.latitude or B.latitude it will not make much difference */

Hope this will help...


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...